Now let us take a look at the change in entropy of a Carnot engine and its heat reservoirs for one full cycle. The hot reservoir has a loss of entropy
ΔSh=−Qh/Thsize 12{ΔS rSub { size 8{h} } = - Q rSub { size 8{h} } /T rSub { size 8{h} } } {} , because heat transfer occurs out of it (remember that when heat transfers out, then
Qsize 12{Q} {} has a negative sign). The cold reservoir has a gain of entropy
ΔSc=Qc/Tcsize 12{ΔS rSub { size 8{c} } =Q rSub { size 8{c} } /T rSub { size 8{c} } } {} , because heat transfer occurs into it. (We assume the reservoirs are sufficiently large that their temperatures are constant.) So the total change in entropy is

This result, which has general validity, means that
the total change in entropy for a system in any reversible process is zero.

The entropy of various parts of the system may change, but the total change is zero. Furthermore, the system does not affect the entropy of its surroundings, since heat transfer between them does not occur. Thus the reversible process changes neither the total entropy of the system nor the entropy of its surroundings. Sometimes this is stated as follows:
Reversible processes do not affect the total entropy of the universe. Real processes are not reversible, though, and they do change total entropy. We can, however, use hypothetical reversible processes to determine the value of entropy in real, irreversible processes. The following example illustrates this point.

How can we calculate the change in entropy for an irreversible process when
ΔStot=ΔSh+ΔScsize 12{ΔS rSub { size 8{"tot"} } =ΔS rSub { size 8{h} } +ΔS rSub { size 8{c} } } {} is valid only for reversible processes? Remember that the total change in entropy of the hot and cold reservoirs will be the same whether a reversible or irreversible process is involved in heat transfer from hot to cold. So we can calculate the change in entropy of the hot reservoir for a hypothetical reversible process in which 4000 J of heat transfer occurs from it; then we do the same for a hypothetical reversible process in which 4000 J of heat transfer occurs to the cold reservoir. This produces the same changes in the hot and cold reservoirs that would occur if the heat transfer were allowed to occur irreversibly between them, and so it also produces the same changes in entropy.

There is an
increase in entropy for the system of two heat reservoirs undergoing this irreversible heat transfer. We will see that this means there is a loss of ability to do work with this transferred energy. Entropy has increased, and energy has become unavailable to do work.

Questions & Answers

How is the de Broglie wavelength of electrons related to the quantization of their orbits in atoms and molecules?

acleration is vectr quatity it is found in a spefied direction and it is product of displcemnt

bhat

its a scalar quantity

Paul

velocity is speed and direction. since velocity is a part of acceleration that makes acceleration a vector quantity. an example of this is centripetal acceleration. when you're moving in a circular patter at a constant speed, you are still accelerating because your direction is constantly changing.

Josh

acceleration is a vector quantity. As explained by Josh Thompson, even in circular motion, bodies undergoing circular motion only accelerate because on the constantly changing direction of their constant speed. also retardation and acceleration are differentiated by virtue of their direction in

fitzgerald

respect to prevailing force

fitzgerald

What is the difference between impulse and momentum?

Manyo

Momentum is the product of the mass of a body and the change in velocity of its motion.
ie P=m(v-u)/t (SI unit is kgm/s). it is literally the impact of collision from a moving body.
While
Impulse is the product of momentum and time.
I = Pt (SI unit is kgm) or it is literally the change in momentum

fitzgerald

Or I = m(v-u)

fitzgerald

the tendency of a body to maintain it's inertia motion is called momentum( I believe you know what inertia means) so for a body to be in momentum it will be really hard to stop such body or object..... this is where impulse comes in.. the force applied to stop the momentum of such body is impulse..

what impulse is given to an a-particle of mass 6.7*10^-27 kg if it is ejected from a stationary nucleus at a speed of 3.2*10^-6ms²? what average force is needed if it is ejected in approximately 10^-8 s?